We consider equation (REF ) for \(k=2\)  with the test function \(\varphi (t,x,y)= \eta (t,y-x)\)  (for simplicity replacing \(x_1\)  and \(x_2\)  by \(x\)  and \(y\)  and writing in component form \(\eta = (\eta ^{ij})_{ij}\) ),
\(\begin{split}&\sum _{ij}\int _0^T\!\!\int _D\!\int _D\int _{U^2}\xi _1^i \xi _2^j\frac{\partial \eta ^{ij}}{\partial t}(t,y-x)\,d\nu _{t,x,y}^2(\xi ) \,dx\,dy \,dt\\&\quad +\sum _{ij}\int _D\!\int _D\int _{U^2}\!\!\xi _1^i \xi _2^j\eta ^{ij}(0,y-x)\,d\nu _{0,x,y}^2(\xi ) \,dx \,dy \\&\quad +\sum _{ijk}\int _0^T\!\!\int _D\!\int _D\int _{U^2}\xi _1^i\xi _1^k\xi _2^j\partial _{x^k}\eta ^{ij}(t,y-x)\,d\nu _{t,x,y}^2(\xi ) \,dx\,dy \,dt\\&\quad +\sum _{ijk}\int _0^T\!\!\int _D\!\int _D\int _{U^2}\xi _1^i\xi _2^j\xi _2^k\partial _{y^k}\eta ^{ij}(t,y-x)\,d\nu _{t,x,y}^2(\xi ) \,dx\,dy \,dt\\&=-\varepsilon \sum _{ij}\int _0^T\!\!\int _D\!\int _D\int _{U^2}\!\!\xi _1^i \xi _2^j(\Delta _{x}+\Delta _y)\eta ^{ij}(t,y-x)\,d\nu _{t,x,y}^2(\xi ) \,dx \,dy \,dt\\&\quad +\sum _{ijk}\int _0^T\!\!\int _D\!\int _D\int _{U^2}\xi _1^i\xi _1^k\xi _2^j\partial _{x^k}\partial _{x^i}\psi ^j_{\eta ,1}(t,y-x)\,d\nu _{t,x,y}^2(\xi ) \,dx \,dy \,dt\\&\quad +\sum _{ijk}\int _0^T\!\!\int _D\!\int _D\int _{U^2}\xi _1^i\xi _2^j\xi _2^k\partial _{y^k}\partial _{y^j}\psi ^i_{\eta ,2}(t,y-x)\,d\nu _{t,x,y}^2(\xi ) \,dx \,dy \,dt\end{split}\) 
Since \(\psi _{\eta ,1}\)  solves \(\Delta _x\psi _{\eta ,1}=\operatorname{div}_x\eta \)  and \(\psi _{\eta ,2}\)  solves \(\Delta _y\psi _{\eta ,2}=\operatorname{div}_y\eta =-\operatorname{div}_x\eta \) , we have \(\psi _{\eta ,1}= (\Delta _x)^{-1}\operatorname{div}_x\eta \)  and \(\psi _{\eta ,2}=-(\Delta _y)^{-1}\operatorname{div}_x\eta =-(\Delta _x)^{-1}\operatorname{div}_x\eta \)  and so \(\psi _\eta \psi _{\eta ,1}=-\psi _{\eta ,2}\)  (up to additive constants). Using this and changing the integration variables to \(x\)  and \(hy-x\) , we obtain
\(\begin{split}&\sum _{ij}\int _0^T\int _D\!\!\int _D\int _{U^2}\xi _1^i \xi _2^j\frac{\partial \eta ^{ij}}{\partial t}(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx\,dh \,dt\\&\quad +\sum _{ij}\int _D\!\!\int _D\int _{U^2}\!\!\xi _1^i \xi _2^j\eta ^{ij}(0,h)\,d\nu _{0,x,x+h}^2(\xi ) \,dx \,dh \\&\quad -\sum _{ijk}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\xi _1^i\xi _1^k\xi _2^j\partial _{h^k}\eta ^{ij}(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx\,dh \,dt\\&\quad +\sum _{ijk}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\xi _1^i\xi _2^j\xi _2^k\partial _{h^k}\eta ^{ij}(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx\,dh \,dt\\&=-2\varepsilon \sum _{ij}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\!\!\xi _1^i \xi _2^j\Delta _h\eta ^{ij}(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx \,dh \,dt\\&\quad +\sum _{ijk}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\xi _1^i\xi _1^k\xi _2^j\partial _{h^k}\partial _{h^i}\psi ^j_{\eta }(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx \,dh \,dt\\&\quad -\sum _{ijk}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\xi _1^i\xi _2^j\xi _2^k\partial _{h^k}\partial _{h^j}\psi ^i_{\eta }(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx \,dh \,dt\end{split}\) 
where \(\psi _\eta = \psi _{\eta ,1}\) .
The cubic terms can be rewritten using the following simple fact (which can also be found in Frisch [1]}, and in a similar, weak form in [2]}):
\(\begin{split}&\quad -2\sum _{ijk}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\xi _1^i\xi _2^j(\xi _1^k-\xi ^k_2)\partial _{h^k}\eta ^{ij}(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx\,dh \,dt\\&=\sum _{ijk}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}(\xi _1^i-\xi ^i_2)(\xi _1^j-\xi _2^j)(\xi _1^k-\xi ^k_2)\partial _{h^k}\eta ^{ij}(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx\,dh \,dt.\end{split}\) 
The proof of this is postponed to the end of this proof. Using this, we can rewrite (REF ) as
\(\begin{split}&\quad \sum _{ij}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\!\!\xi _1^i \xi _2^j\frac{\partial \eta ^{ij}}{\partial t}(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx\,dh \,dt\\&\quad +\sum _{ij}\int _D\!\!\int _D\int _{U^2}\!\!\xi _1^i \xi _2^j\eta ^{ij}(0,h)\,d\nu _{0,x,x+h}^2(\xi ) \,dx \,dh \\&\quad -\frac{1}{2}\sum _{ijk}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}(\xi _1^i-\xi ^i_2)(\xi _1^j-\xi _2^j)(\xi _1^k-\xi ^k_2)\partial _{h^k}\eta ^{ij}(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx\,dh \,dt\\&=-2\varepsilon \sum _{ij}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\!\!\xi _1^i \xi _2^j\Delta _h\eta ^{ij}(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx \,dh \,dt\\&\quad +\sum _{ijk}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\xi _1^i\xi _1^k\xi _2^j\partial _{h^k}\partial _{h^i}\psi ^j_{\eta }(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx \,dh \,dt\\&\quad -\sum _{ijk}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\xi _1^i\xi _2^j\xi _2^k\partial _{h^k}\partial _{h^j}\psi ^i_{\eta }(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx \,dh \,dt.\end{split}\) 
Since
\(&\quad \sum _{ij}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\!\!\xi _1^i \xi _1^j\Delta _h\eta ^{ij}(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx \,dh \,dt\\&= \sum _{ij}\int _0^T\!\!\int _{D}\!\int _{U}\!\!\xi _1^i \xi _1^j \,d\nu _{t,x}^1(\xi ) \,dx \int _{D} \Delta _h\eta ^{ij}(t,h)\,dh \,dt=0,\) 

we can rewrite
\(&\quad -2\varepsilon \sum _{ij}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\!\!\xi _1^i \xi _2^j\Delta _h\eta ^{ij}(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx \,dh \,dt\\&=\varepsilon \sum _{ij}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\!\!(\xi _1^i-\xi _2^i)(\xi _1^j- \xi _2^j)\Delta _h\eta ^{ij}(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx \,dh \,dt.\) 

Moreover, we have for symmetric, smooth and compactly supported rank 2 tensors \(\eta \)  of the form
\(\eta (t,h)= \omega _1(t,|h|) \mathbf {I}+ \omega _2(t,|h|)\hat{h}\otimes \hat{h},\) 
with \(\psi ^i_\eta = -(\Delta _h)^{-1}\operatorname{div}_h\eta ^{i,\cdot }\) ,
\(\begin{split}&\sum _{ijk}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\xi _1^i\xi _1^k\xi _2^j\partial _{h^k}\partial _{h^i}\psi ^j_{\eta }(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx \,dh \,dt = 0,\\&\sum _{ijk}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\xi _1^i\xi _2^j\xi _2^k\partial _{h^k}\partial _{h^j}\psi ^i_{\eta }(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx \,dh \,dt=0,\end{split}\) 
whose proof is postponed to the end of this proof. Using this, (REF ) becomes
\(\begin{split}&\sum _{ij}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\!\!\xi _1^i \xi _2^j\frac{\partial \eta ^{ij}}{\partial t}(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx\,dh \,dt\\&\quad +\sum _{ij}\int _D\!\!\int _D\int _{U^2}\!\!\xi _1^i \xi _2^j\eta ^{ij}(0,h)\,d\nu _{0,x,x+h}^2(\xi ) \,dx \,dh \\&\quad -\frac{1}{2}\sum _{ijk}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}(\xi _1^i-\xi ^i_2)(\xi _1^j-\xi _2^j)(\xi _1^k-\xi ^k_2)\partial _{h^k}\eta ^{ij}(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx\,dh \,dt\\&=\varepsilon \sum _{ij}\int _0^T\!\!\int _D\!\!\int _D\int _{U^2}\!\!(\xi _1^i-\xi _2^i)(\xi _1^j- \xi _2^j)\Delta _h\eta ^{ij}(t,h)\,d\nu _{t,x,x+h}^2(\xi ) \,dx \,dh \,dt.\end{split}\) 
Let \(\theta _\delta \)  be a sequence of smooth, uniformly bounded functions with the property that \(\theta _\delta \rightarrow \mathbf {1}_{(0,\tau ]}(t)\)  for every \(t\)  as \(\delta \rightarrow 0\) . If we now use a test function
\(\eta (t,h)= \sigma (h)\theta _\delta (t),\) 
where \(\sigma \)  is of the form (REF ), then we can use the weak continuity in time of the moments \(\int _{U^k}\xi _1\otimes \dots \otimes \xi _k \,d\nu ^k_{t,x}(\xi )\)  to obtain for any \(\tau >0\) , as \(\delta \rightarrow 0\) ,
\(\begin{split}&\quad -\sum _{ij}\int _D\!\!\int _D\int _{U^2}\!\!\xi _1^i \xi _2^j\,d\nu _{\tau ,x,x+h}^2(\xi ) \,dx \sigma ^{ij}(h) \,dh +\sum _{ij}\int _D\!\!\int _D\int _{U^2}\!\!\xi _1^i \xi _2^j\,d\nu _{0,x,x+h}^2(\xi ) \,dx \sigma ^{ij}(h) \,dh \\&\quad -\frac{1}{2}\sum _{ijk}\int _0^\tau \!\!\int _D\!\!\int _D\int _{U^2}(\xi _1^i-\xi ^i_2)(\xi _1^j-\xi _2^j)(\xi _1^k-\xi ^k_2)\,d\nu _{t,x,x+h}^2(\xi ) \,dx \partial _{h^k}\sigma ^{ij}(h)\,dh \,dt\\&=\varepsilon \sum _{ij}\int _0^\tau \!\!\int _D\!\!\int _D\int _{U^2}\!\!(\xi _1^i-\xi _2^i)(\xi _1^j- \xi _2^j)\,d\nu _{t,x,x+h}^2(\xi ) \,dx\Delta _h\sigma ^{ij}(h) \,dh \,dt.\end{split}\) 
